Let $X$ be a connected $CW$-complex, such $\pi_1(X)$ is torsion-free and $H_k(X,\mathbb Z) = 0$ for all $k \geq N$ and some $N \in \mathbb N$. Then

$(1)$ Does it follow that $X$ is homotopy-equivalent to its $N$-skeleton, i.e $X \simeq X^{(N)}$ ?

$(2)$ If $(1)$ is false, does it follow that $X \simeq X^{(k)}$ for some $k \in \mathbb N$ ?

$(3)$ If $(2)$ is also false, at the very least, does it follow that $X \simeq Y$ for some finite-dimensional $CW$-complex $Y$ ?

It is clear to me that the question has a negative answer when $H_k$ is replaced by $\pi_k$ (take a $CW$-model of $BG$ for a group $G$ containing elements of finite order). I am also very certain that this question has been asked before and I just couldn't find the corresponding thread, so I have no problem with this question being marked as duplicate.

Edit: One could then possibly find a *finitely dominated* $CW$-complex $X$, with $\pi_1(X)$ not torsion-free, such that its *Wall finiteness obstruction* element $w(X) \in \widetilde{K_0}(\mathbb Z[\pi_1(X)])$ is non-trivial. Any finitely dominated $CW$-complex has finite homological-dimension (so it satisfies the assumptions), but if its finiteness obstruction doesn't vanish, it doesn't have the homotopy type of a finite complex.

Conversely, if $X$ is finitely-dominated and $w(X) = 0$, then it has the homotopy type of a finite-dimensional complex.

In particular, this must be the case if $\widetilde{K_0}(\mathbb Z[\pi_1(X)]) =0$. It is conjectured that $\widetilde{K_0}(\mathbb Z[G]) = 0$ if $G$ is torsion-free, hence the assumption in the statement of the question.

**Edit 2: Mark has given an example of a $CW$-complex $X$ with finite homological dimension, but infinite cohomological dimension (and thus, has given a negative answer to my original questions) . It would be still interesting to know if there exist counter-examples $X$, such that for any $\pi_1(X)$-module $M$, both the homology $H_*(X,M)$ and the cohomology $H^*(X,M)$ have groups only in finitely many dimensions.**